All categories

3 years ago

The barometric pressure in breckenridge, colorado (elevation 9600 feet) is 580 mm hg. how many atmospheres is this?

Physics

1 answer:

Maksim231197 [3]3 years ago

7 0

1 atmospheric pressure = 760.0 mm Hg

Thus 580 mm Hg = (580 mm Hg/(760 mm Hg/atm))

= 0.763 atm

You might be interested in

A dog walks 12 meters to the east and then 16 meters back to the west for this motion what is the distance moved What is the mag

FinnZ [79.3K]

The distance is 28 meters, and the displacement is -4.

For the distance it would be 12 + 16 = 28.

For the displacement it would be 12 - 16 = -4.

would really appreciate a brainliest! Hope this helped!

6 0

3 years ago

y en el número racional (Q) se puede escribir como conciente de dos números enteros (z)o como un desimal periódico.Espresa el 1/

umka21 [38]

Answer:

Sorry I don't understand this language I'm sorry

6 0

2 years ago

An atom that has fewer neutrons than protons and more electrons than protons is a ion

riadik2000 [5.3K]

i think it is an ion

6 0

3 years ago

Why doesn’t she love me

o-na [289]

I bet she does just give her tule work on yourself

7 0

3 years ago

Read 2 more answers

One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th

lozanna [386]

Answer:

$k=105.0359\times 10^4\,W.m^{-1}.K^{-1}$

Explanation:

Given:

temperature at the hotter end, $T_H=100^{\circ}C$

temperature at the cooler end, $T_C=0^{\circ}C$

length of rod through which the heat travels, $dx=0.7\,m$

cross-sectional area of rod, $A=1.1\times 10^{-4}\,cm^2$

mass of ice melted at zero degree Celsius, $m=8.7\times 10^{-3}\,kg$

time taken for the melting of ice, $t=15\times60=900\,s$

thermal conductivity k=?

By Fourier's Law of conduction we have:

$\dot{Q}=k.A.\frac{dT}{dx}$ ......................................(1)

where:

$\dot{Q}$ =rate of heat transfer

dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice, $L = 334000\,J.kg^{-1}$

$Q=m.L$

$Q=8.7\times 10^{-3}\times 334000$

$Q=2905.8\,J$

Now the heat rate:

$\dot{Q}=\frac{Q}{t}$

$\dot{Q}=\frac{2905.8}{900}$

$\dot{Q}=3.2287\,W$

Now using eq,(1)

$3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}$

$k=205.4606\,W.m^{-1}.K^{-1}$

8 0

3 years ago